An aperiodic set of 11 Wang tiles - Sciencesconf.org · 2016-06-14 · De nitions and history...

De�nitions and historyGeneration of set with at most 10 tiles

Aperiodic set of 11 tiles

An aperiodic set of 11 Wang tiles

Emmanuel Jeandel 1 Michaël Rao 2

1LORIA - Nancy

2LIP - Lyon

E. Jeandel, M. Rao An aperiodic set of 11 Wang tiles



A Wang tile is a square tile with a color on each border

Given a set of Wang tiles, one try to tile the plane with copies of

tiles in the set s.t. two adjacent sides have the same color

(No rotations !)




A tiling of the plane is periodic if there is a translation vector which

does not change the tiling

A tile set is periodic if there is a periodic tiling of the plane with

this set




A set is periodic if and only if there is a tiling with 2 (not colinear)

translation vectors




A set is �nite if there is no tiling of the plane with this set

A set is aperiodic if it tiles the plane, but no tiling is periodic

Conjecture (Wang 1961)

Every set is either �nite or periodic

False:

Theorem (Berger 1966)

It exists an aperiodic set of Wang tiles




History

Berger : 20426 tiles in 1966 (lowered down later to 104)

Knuth : 92 tiles in 1968

Robinson : 56 tiles in 1971

Ammann : 16 tiles in 1971

Grunbaum : 24 tiles in 1987

Kari and Culik : 14 tiles, then 13 tiles in 1996

Here : 11 tiles (the fewest possible)




�Kari-Culik� tile set

Theorem (Kari-Culik 1996)

The following set (13 tiles) is aperiodic

0 11

20 21

11 21

21 00

12 00

22 10

1

0′ 0′0′

00′ 0′2

10′

12

1

00′

12

1

0′12

12

0′

0

12

12

2

1

12 0′1

1




New results

Theorem

Every set with at most 10 Wang tiles is either �nite or periodic

Theorem

There is a set with 11 Wang tiles which is aperiodic




Transducer viewPowerEnumeration and results

Transducer

A set of Wang tiles can be seen as a transducer

A transducer is a �nite automaton where each transition has an

input letter and an output letter

T = (H,V ,T ) where T ⊆ H2 × V 2

We note wT w ′ if the transducer T writes w ′ when it reads w

(Transducer on Σ = Automaton on Σ2)





�Kari-Culik� tile set

1

02 0′

12

1|2

1|1

1|2

0|1

0|2

0|10′|02|1

1|01|0′

0′|02|1

1|1

0 11

20 21

11 21

21 00

12 00

22 10

1

0′ 0′0′

00′ 0′2

10′

12

1

00′

12

1

0′12

12

0′

0

12

12

2

1

12 0′1

1

Figure: Kari-Culik tile set.





Simpli�cation

If T has a transition a between two strongly connected

components, then T is �nite (resp. periodic) if and only if T \ {a}is �nite (resp. periodic)

Let s(T ) be the union of strongly connected components of T

T is �nite (resp. periodic, aperiodic) if and only if s(T ) is �nite

(resp. periodic, aperiodic)





Composition and power

Let T = (H,V ,T ) and T ′ = (H ′,V ,T ) be two transducers

Then T ◦ T ′ = (H × H ′,V ,T ′′) where:

T ′′ = {((w ,w ′), (e, e ′), s, n′) : (w , e, s, x) ∈ T , (w ′, e ′, x , n′)}

T k = T k−1 ◦ T





Power

Proposition

There is k ∈ N s.t. s(T k) is empty i� T is �nite

Proposition

There is k ∈ N s.t. there is a bi-in�nite word w such that wT kwi� T is periodic





Enumeration (I)

To enumerate all sets with n tiles, we compute all oriented graphs

with n arrows (with loops and multiple arrows)

For every pair of graphs G and G ′, we try every n! bijectionsbetween the arrows of G and G ′

We only consider graphs without arrows between two strongly

connected components.

n nb. graphs

8 2518

9 13277

10 77810

11 493787





Enumeration (II)

For every generated set T , we compute s(T k) until:

s(T k) is empty → the set is �nite

∃w s.t. ws(T k)w → is periodic

The computer run out of memory → the computer cannot

conclude...

Optimizations :

Cut branches in the exploration of n! bijections

Make tests on T and T tr on the same time

Use (sometimes) bi-simulation to simplify transducers





Result (n ≤ 10)

Theorem

Every set of n Wang tile, n ≤ 10, is �nite or periodic

∼ 4 days on ∼ 100 cores

Only one problematic case





The only problematic case with 10 tiles

1 0

1|2 1|21|3, 0|1

2|3, 1|1

2/3 0′

3|1 3|11|1

2|0

�Kari-Culik� type ×2, ×13 . We have to use compactness to show

that this tile-set is �niteE. Jeandel, M. Rao An aperiodic set of 11 Wang tiles



From T to TDFrom TD to T0,T1,T2From Tn,Tn+1,Tn+2 to Tn+1,Tn+2,Tn+3


0

1

23

1|02|1

2|2

4|2

2|3

1|1

1|12|2

3|1 1|40|2

0 01

00 32

11 02

21 14

21 32

33 01

1

3 11

13 12

23 33

12 21

42 20

2





Proof (squetch.)

Theorem

T is aperiodic

Ideas:

It is the union if two transducers T0 and T1 (as for Kari-Culik)

In a tiling by T , we can merge layers into T10000 and T1000We get a new transducer TD with 28 transitions, which is the unionof Ta ' T10000 and Tb ' T1000We de�ne the family Tn, with Tn+3 = Tn+1 ◦ Tn ◦ Tn+1

We show Tb = T0, Taa = T1, Tbab = T2

The only admissible vertical word for TD is the Fibonacci word





T is the union of T0 and T1, with (resp.) 9 and 2 tiles

For w ∈ {0, 1}∗ \ {ε}, let Tw = Tw [1] ◦ Tw [2] ◦ . . . Tw [|w |]

Fact

s(T11), s(T101), s(T1001) and s(T00000) are empty

If t is a tiling by T , then there exists a bi-in�nite word

w ∈ {1000, 10000}Z s.t. t(x , y) ∈ T (Tw [y ])

Let TA = s(T1000 ∪ T10000)

There is a bijection between tilings by T and tilings by TA





20330 21030

21033

21100

21103

21113

21130

21300

21310

21311

21330

23100

23300

23310

1|0

1|11|0

1|1

1|1

1|0

1|11|1

1|01|0

1|1

1|21|3

1|1

0|00|0

0|00|0

0|0

0|0

0|1

20302033

2100

2103

2110

2111

2113

2130

2131

2133

2300

2310

2311

2330

2331

1|0

1|1

1|1

1|1

1|2

1|3

1|1

1|1

1|2

1|3

1|2

0|00|1

0|00|1

0|1

0|0

0|1

0|1

0|0

0|1

0|1

0|00|1

0|2

0|3

0|1

0|2

0|3

Figure: TA, the union of s(T10000) (top) and s(T1000) (bottom).





Elimination of transitions with 2, 3 or 4

20330 21030

21033

21103

21113

21130

21300

21310

21330

23100

23300

23310

1|0

1|11|0

1|11|11|1

1|01|0

1|1

1|1

0|00|0

0|00|0

0|0

0|0

0|1

20302033

2103

2113

2130

2133

2300

2310

23301|1

1|1

1|1

0|00|1

0|00|1

0|1

0|1

0|0

0|10|0

0|10|1

Figure: TB = s(s(TAtr)tr

).





Simpli�cation by bi-simulation

a

b

c

d e f

gh

i j

1|0

1|1 1|11|1

1|01|0

1|1

1|1

0|00|0

0|0

0|0

0|0

0|1

K

L

M N O

P

Q

R1|1

1|1

1|1

0|00|1

0|1

0|1

0|0

0|10|0

0|10|1

Figure: TC , �simpli�cation� of TB .





Proposition

Let (wi )i∈Z be a bi-in�nite sequence of bi-in�nite words s.t.

wiTCwi+1 for every i ∈ Z.Then for every i ∈ Z, wi is (010, 101)-free

This follow from the fact than s((TC tr)3) does not contains the

state 010, nor the state 101

Every tiling by TC is in bijection with a tiling by TD





a

b

c

d e f

gh

i j

1|0

1|1 1|11|1

1|01|0

1|1

1|1

0|00|0

0|0

0|0

0|0

0|1

K

L M

M'

N O

O'

P

QR

1|1

1|1

1|1

0|0

0|1

0|1

0|1

0|00|10|0

0|10|1

0|01|1

Figure: TD , the union of Ta (top) and Tb (bottom)





Case of Tb

In Tb, every long enough path passes through �N�

Thus Tb is equivalent to:

N

00000 |1001100000000 |1110001100111000 |1111111100110 |111110000011000|11100111110010 |1011001000 |1110110000010 |11100110011000 |1011111





Case of TbIn Tb, every long enough path passes through �N�

Thus Tb is equivalent to:

N

00000 |1001100000000 |1110001100111000 |1111111100110 |111110000011000|1110011111

Tb is equivalent to T0.

ε|ε

05 |(100)1205+3 |13(000)1202(111)03|15+302(110) |12+305(110)02|12(100)15





Case of Taa

ba

ch

cj

dc

eb

gb

111|000

111|000

11|1111|00

1|0

11111100|11000000

1|0

1111|0111

0001|000000|00

Figure: s(Taa)





In s(Taa), every long path passes through �eb�. It is equivalent to:

eb

11111111 |110000001111111111111 |00000111000001110001111111 |000000000000011110011 |000000001111111110011111|0001100000000000

Taa is equivalent to T1

15−3|05−3

13+3 |(110)0318+3 |05(111)0313(000)15|08+3)13(100) |03+318(100)13|03(110)08





Case of Tbab

LaO

MbK

MbR

MdP

NcL

PbN

QcO

RcO 1|1

1|1

000000011|001111111

0000|1100

0000|1111

0|1

0|0

0|1

0000000|1111111

0|0

0000|1111

1100|1111

Figure: s(Tbab)





Every long path passes through �MdP�. Thus it is equivalent to :

MdP

0000000000000 |1111111001111000000000000000000000 |111111111111110001111000000000001110000000 |1111111111111111111110000000000011 |111111111111100000000000000000001100000|11111111111100111111111111

If we shift the input (3 times) and the output (6 times), we get:

0000000000000 |1001111111111000000000000000000000 |111111110001111111111000000001110000000000 |1111111111111111111110000000011000 |111111111111100000000000000001100000000|11111100111111111111111111

Tbab is equivalent to T2





Fact

s(Tbb), s(Taaa) and s(Tbabab) are empty

If t is a tiling by TD , then there is a bi-in�nite word

w ∈ {b, aa, bab}Z s.t. t(x , y) ∈ T (Tw [y ])

That is, the tilings with TD are images of the tilings by

Tb ∪ Taa ∪ Tbab ' T0 ∪ T1 ∪ T2.





One supposes than n is even. (The odd case is similar.)

One supposes that Tn ∪ Tn+1 ∪ Tn+2 tiles the plane

We show that Tn+1 ∪ Tn+2 ∪ Tn+3 tiles, and that:

Tn+3 ' Tn+1 ◦ Tn ◦ Tn+1.

Tn is surrounded by Tn+1

(output of Tn+1 has more 0's than 1's, Tn and output Tn+2 have

more 1's than 0's)





Transitions of Tn:

α γ

δ β

ε ω

Transitions of Tn+1:

A C

D B

E O





Let's take Tn. (We forget α.)

Lemma

The following words cannot appear:

γω,γγ,γβ, βω, ββ, βεβ, γεβ, βδεβ, γδεβ

ωδ, δδ, εδ, ωε, εε, εβε, εβδ, εβγε, εβγδ

ω





Lemma

Every in�nite path in the transducer Tn can be seen as an in�nite

path in the following transducer:

0 5

γδ

β, ε, βδγε, βγε, βδε





Lemma

Every in�nite path in the transducer Tn can be seen as an in�nite

path in the following transducer:

0 5

γαδ

αβα, αεα, αβαδαγαεα, αβαγαεα, αβαδαεα





γδ → A′

γ α δ

A C

D A

ε→ B′

α ε α

A B

B A





β → E′:

α β α

A E

E A

βγε→ C′:

α β α δ α ε α

A C A B

E A C A





βδε→ D′:

α β α γ α ε α

A D A B

E A D A

βδγε→ O′:

α β α δ α γ α ε αA C A E A B

E A B A D A





It remains to show that one cannot have a stack of layers

Tn+1,Tn,Tn+1,Tn,Tn+1

We can merge layers of a tiling with Tn ∪ Tn+1 ∪ Tn+2 by:

1 Tn+1

2 Tn+2

3 Tn+1TnTn+1 ' Tn+3





Tn ' Tun where:

u0 = b, u1 = aa, u2 = bab

and

un+3 = un+1unun+1

(un)n≥0 = (b, aa, bab, aabaa, babaabab, . . .)

is a sequence of factors of the Fibonacci word

(the �singular factors�)





Fibonacci Word

Aperiodic word

wf = abaababaabaababaababaabaababaabaababaababaa...

Fixed point of the morphism a→ ab, b → a

v0 = a, v1 = ab and vn+2 = vn+1vnvi converges to wf

(vn)n≥0 = (a, ab, aba, abaab, abaababa, abaababaabaab, . . .)

(un)n≥0 = (b, aa, bab, aabaa, babaabab, aabaababaabaa, . . .)E. Jeandel, M. Rao An aperiodic set of 11 Wang tiles




Aperiodic set with 11 tiles

0

1

23

1|02|1

2|2

4|2

2|3

1|1

1|12|2

3|1 1|40|2

0 01

00 32

11 02

21 14

21 32

33 01

1

3 11

13 12

23 33

12 21

42 20

2





Aperiodic set with 11 tiles and 4 colors

0

1

23

1|02|1

2|2

0|2

2|3

1|1

1|12|2

3|1 1|00|2

0 01

00 32

11 02

21 10

21 32

33 01

1

3 11

13 12

23 33

12 21

02 20

2





Open question 1 : Another aperiodic set ?

Tiles sets with 11 tiles:

2 aperiodic (and 1 other probably very close)

23 others �candidates�

9 of �Kari-Culik� type (and probably �nite)

14 not �Kari-Culik�

1 strange (interesting) candidate:

4 00

00 10

00 11

11 21

12 20

22 21

2

2 32

22 42

13 42

04 40

24 41

0





Open question 2 : �proof from the book� ?

If we look at densities of 1 on each line on an in�nite tiling, one

transducer add ϕ− 1 and the other add ϕ− 2.

→ �additive� Kari-Culik ?


An aperiodic set of 11 Wang tiles - Sciencesconf.org · 2016-06-14 · De nitions and history...

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